Learning material for this module

• James et al (2013): An Introduction to Statistical Learning. Chapter 2.2.3 and 4.

We need more statistical theory than is presented in the textbook, which you find in this module page.

Topics in this module

Part A: Introduction to classification, and modelling class densities

• Aim of this module
• What is classification and what is discrimination?
• Loss function and the Bayes classifier
• Modelling class densities
  • Linear discriminant analysis
  • Quadratic discriminant analysis
  • Naive Bayes (optional)
• Modelling posterior probabilities
  • KNN-classifier

Part B: Modelling posterior probabilites, ROC/AUC and comparisions

• Modelling posterior probabilities (cont.)
  • Linear regression for classification problems
  • Logistic regression
• Sensitivity, specificity, ROC and AUC
• Comparisons
• Extensions

Part A

What will you learn?

• What is classification and discrimination?
• What is the Bayes classifier and the Bayes risk?
• What is the sampling paradigm and what is modelled then?
• Linear discriminant analysis: model, method, results.
• Quadratic discriminant analysis: model, method, results.
• Naive Bayes - when and why?
• That is the diagnostic paradigm and what is modelled then?
• KNN - majority vote or estimate posterior class probability?

Set-up

Training set: observations (independent pairs) \(\{(x_1, y_1), ..., (x_n, y_n)\}\) where the response variable \(Y\) is qualitative and labelled \(1, 2, ..., K\).

The training set is used to construct the classification rule (or similar).

Test set: observations (independent pairs), same format as the training set.

The test set is used to evaluate the classification rule.

Loss function: The misclassifications are given the loss 1 and the correct classifications loss 0 - this is called 0/1-loss. (Quadratic loss is not used for classification.)

But, what if the cost of misclassification is different for the classes? In a more general set-up we could also take that into account. Read more about this: Decision theoretic framework (sample space, action space, loss function, risk function, Bayes risk, cost function).

Bayes theorem - for discrimination and classification

We have that \(X\) is a continuous random vector and \(Y\) is a univariate categorical random variable.

Bayes theorem \[\begin{align*} P(Y=k \mid {\bf X}={\bf x}) &= \frac{P({\bf X}={\bf x} \cap Y=k)}{f({\bf x})}\\ &= \frac{\pi_k f_k(x)}{\sum_{l=1}^K \pi_l f_l(x)} \end{align*}\]

Here \(f_k(x)\) is the pdf for \(X\) in class \(k\) and \(\pi_k = P(Y=k)\) is the prior probability for class \(k\).

Synthetic example: what is the Bayes error?

Suppose we have observations coming from two classes: {green, orange}, where \[X_{\text{green}}\sim \mathcal{N}(-2, 1.5^2) \text{ and } X_{\text{orange}}\sim \mathcal{N}(2, 1.5^2) \] and that the probability of observing each class is equal.

• Where is the Bayes decision boundary here?
• How can we calculate the Bayes error rate (intuitively- see the graph)?
• What would you estimate the Bayes error rate to be?
• What if someone constructs a classifier and it has a lower error rate than the Bayes error rate - is this possible?

Bayes error: round(0.5*2*pnorm(0,mean=2,sd=1.5),2)=0.09

Bayes decision rule - two paradigms

The most popular approach to classication is to use the Bayes decision rule with the 0/1 loss function= classify to the class with the largest \(P(Y=k\mid X=x_0)\).

Two approaches:

• The diagnostic paradigm: We focus on directly estimating the posterior distribution for the classes \(P(Y=k \mid X=x)\).

• The sampling paradigm: There focus is on estimating the prior probabilities \(\pi_k\) for the classes and the class conditional distributions \(f_k(x)\). We classify to the class with the maximal product \(\pi_k f_k(x)\).

We first look at the sampling paradigm - and then we need to model the pdf for each class. Popular: the multivariate normal distribution!

Univariate normal class distributions - the role of the prior

Suppose (again) we have observations coming from two classes: {green, orange}, where \[X_{\text{green}}\sim \mathcal{N}(-2, 1.5^2) \text{ and } X_{\text{orange}}\sim \mathcal{N}(2, 1.5^2) \]

In the figure below we have specified the prior probabilities to be equal, \(\pi_1 = \pi_2 = 0.5\) and have plotted \(\pi_k f_k(x)\) for the two classes.

The decision boundary is where the point of intersection of the two lines is, because here \(\pi_1 f_1(x)=\pi_2 f_2(x)\). Thus all points to left of the decision boundary will be classified as green and similarly, all points to the right of the decision boundary will be classified as orange.

We now specify different priors, such that \(\pi_1 = 0.3\) and \(\pi_2 = 0.7\). We see that this has affected the decision boundary, which now has shifted to the left.

Linear discriminant analysis (LDA)

Using Linear discriminant analysis we assume the class conditional distributions are normal (Gaussian).

Parameter estimators

In real life situations we will not know the prior probabilities, the class mean or standard deviation=need parameter estimators.

• Prior probability for class \(k\) is (often) estimated by taking the fraction of observations \(n_k\) (out of \(n\)) coming from class \(k\): \(\hat{\pi}_k = \frac{n_k}{n}.\)

• The mean value for class \(k\) is simply the sample mean of all observations from class \(k\): \[\hat{\mu}_k = \frac{1}{n_k}\sum_{i:y_i=k} x_i.\]

• The standard deviation: sample standard deviation across all classes: \[\hat{\sigma}^2=\frac{1}{n-K}\sum_{k=1}^K \sum_{i: y_i=k} (x_i-\hat{\mu}_k)^2 = \sum_{k=1}^K \frac{n_k - 1}{n - K} \cdot \hat{\sigma}_k^2.\] \(\hat{\sigma}_k\): estimated standard deviation of all observations from class \(k\).

How would the estimation affect our misclassification rate?

If \(\mu_1=-2\), \(\mu_2=2\), \(\sigma=1.5\), \(\pi_1=\pi_2=0.5\) we found that the class boundary is at 0 and the Bayes error is round(2*0.5*pnorm(0,2,1.5))=0.09.

But, in a real life situation

• we estimate the class boundary
• and we do not know the true distribution for the classes.

How can we then estimate the goodness of our estimator?

1. Use the training set to estimate parameters and class boundary:
2. Use the test set to estimate misclassification rate.

n = 1000
pi1 = pi2 = 0.5
mu1 = -2
mu2 = 2
sigma = 1.5
set.seed(1)
n1train = rbinom(1, n, pi1)
n2train = n - n1train
n1test = rbinom(1, n, pi1)
n2test = n - n1test
train1 = rnorm(n1train, mu1, sigma)
train2 = rnorm(n2train, mu2, sigma)
test1 = rnorm(n1test, mu1, sigma)
test2 = rnorm(n2test, mu2, sigma)
sigma2.1 = var(train1)
sigma2.2 = var(train2)
estsigma2 = ((n1train - 1) * sigma2.1 + (n2train - 1) * sigma2.2)/(n - 
    2)

rule = 0.5 * (mean(train1) + mean(train2)) + estsigma2 * (log(n2train/n) - 
    log(n1train/n))/(mean(train1) - mean(train2))

c((sum(train1 > rule) + sum(train2 < rule))/n, (sum(test1 > rule) + sum(test2 < 
    rule))/n)

## [1] 0.105 0.115

Training error rate

the proportion of mistakes that are made if we apply classifier \(\hat{f}\) to the training observations, i.e. \(\hat{y}_i=\hat{f}(x_i)\). \[\frac{1}{n}\sum_{i=1}^n \text{I}(y_i \neq \hat{y}_i).\] Here I is the indicator function (to give our 0/1 loss) which is defined as: \[\text{I}(a\neq\hat{a}) = \begin{cases} 1 \text{ if } a \neq \hat{a} \\ 0 \text{ else } \end{cases}\] The indicator function counts the number of times our model has made a wrong classification. The training error rate is the fraction of misclassifications made on our training set.

A very low training error rate may imply overfitting.

Test error rate

Here the fraction of misclassifications is calculated when our model is applied on a test set. From what we have learned about regression we can deduct that this gives a better indication of the true performance of the classifier (than the training error). \[\text{Ave}(I(y_0\neq \hat{y}_0))\] where the average is over all the test observations \((x_0,y_0)\).

We assume that a good classifier is a classifier that has a low test error.

The confusion matrix

The confusion matrix is a table that can show the performance of classifier, given that the true values are known.

We can make a confusion matrix from the training or test set, but will in most cases do so for the test set only.

The rows represent the true classes, while the columns represent the predicted classes. Inside the table we have counts (just labelled “correct” and “wrong” below - but should be numbers). The sum of the diagonal is the total number of correct classifications. The sum of all elements off the diagonal is the total number of misclassifications.

The confusion matrix can be obtained in R by using the table function with the predicted classes and the true classes need to be given as function arguments, or directly using the caret package.

We will soon come back to the special case of two classes - where we may think of this as “-” (non disease) and “+” (disease).

Posterior probabilites

Sometimes the probability that an observation comes from a class \(k\) is more interesting than the actual classification itself. These class probabilities can be estimated from the priors and class conditional distributions, or from the discriminant functions: \[\begin{align*}\hat{P}(Y=k | X=x)&= \frac{\hat{\pi}_k \cdot \frac{1}{(2 \pi)^{p/2}|\hat{\boldsymbol{\Sigma}}|^{1/2}} \exp(-\frac{1}{2} ({\bf x}-\hat{\boldsymbol\mu}_k)^T \hat{\boldsymbol{\Sigma}}^{-1} ({\bf x}-\hat{\boldsymbol\mu}_k))} {\sum_{l=1}^K \hat{\pi}_l \frac{1}{(2 \pi)^{p/2}|\hat{\boldsymbol{\Sigma}}|^{1/2}} \exp(-\frac{1}{2} ({\bf x}-\hat{\boldsymbol\mu}_l)^T \hat{\boldsymbol{\Sigma}}^{-1} ({\bf x}-\hat{\boldsymbol\mu}_l))}\\ &= \frac{e^{\hat{\delta}_k(x)}}{\sum_{l=1}^K e^{\hat{\delta}_l(x)}}.\end{align*}\]

Quadratic Discriminant Analysis (QDA)

In quadratic discriminant analysis we assume that the distributions of the classes is multivariate normal (Gaussian), but with covariance matrix \(\boldsymbol{\Sigma}_k\) for each class.

The discriminant functions are now given by: \[\begin{align*} \delta_k(x) &= -\frac{1}{2}(x-\mu_k)^T \boldsymbol{\Sigma}_k^{-1}(x-\mu_k)-\frac{1}{2}\log |\boldsymbol{\Sigma}_k| + \log \pi_k \\ &= -\frac{1}{2} x^T \boldsymbol{\Sigma}_k^{-1}x + x^T \boldsymbol{\Sigma}_k^{-1}\mu_k - \frac{1}{2} \mu_k^T \boldsymbol{\Sigma}_k^{-1}\mu_k - \frac{1}{2}\log |\boldsymbol{\Sigma}_k | + \log \pi_k.\end{align*}\] These decision boundaries are quadratic functions of \(x\).

LDA vs QDA

In QDA we are allowed for the covariance matrices to be different for the classes, but for LDA they are assumed to be the same, so QDA is more flexible than LDA.

Q:

• But, if the covariance matrices in theory are equal - will they not be estimated equal?
• Should we not always prefer QDA to LDA?

A:

Explanation similar to a “Bias-variance trade-off”:

If the assumption of equal covariance matrices is wrong

• then LDA may suffer from high bias for the parameter estimators.
• and QDA is better off. But, for small sample sizes the covariance matrices might be poorly estimated (high variance of estimators).

If the number of covariates is high:

• then QDA requires estimating \(K\cdot p \cdot (p+1)/2\) parameters,
• while LDA only requires \(p\cdot(p+1)/2\).

Therefore, LDA is less flexible than QDA and might therefore have much less variance.

Naive Bayes (optional)

This is method that is popular when \(p\) is large.

In Naive Bayes (Idiot’s Bayes) we assume that each class density is the product of marginal densities - i.e. inputs are conditionally independent in each class. \[f_k(x)=\prod_{j=1}^p f_{kj}(x_j)\] This is generally not true, but it simplifies the estimation dramatically.

The original naive Bayes used univariate normal marginal distributions, but generalizations can be made.

Instead of estimating a full covariance matrix, only the diagonal elements are estimated.

This method often produces good results, even though the joint pdf is not the product of the marginal pdf. This might be because we are not focussing on estimation of class pdfs, but class boundaries.

Example: Classification of iris plants

We will use sepal width and sepal length to build a classificator, here 50 observations from each class available.

attach(iris)
library(class)
library(MASS)
library(ggplot2)
library(dplyr)
kable(head(iris), format = whichformat)

iris0_plot = ggplot(iris, aes(x = Sepal.Width, y = Sepal.Length, color = Species)) + 
    geom_point(size = 2.5)
iris0_plot

# reporting on parameter estimates

grS = cbind(iris[iris$Species == "setosa", 1:2])
grVe = cbind(iris[iris$Species == "versicolor", 1:2])
grVi = cbind(iris[iris$Species == "virginica", 1:2])
nk = dim(grS)[1]
apply(grS, 2, mean)

## Sepal.Length  Sepal.Width 
##        5.006        3.428

cS = cov(grS)
apply(grVe, 2, mean)

## Sepal.Length  Sepal.Width 
##        5.936        2.770

cVe = cov(grVe)
apply(grVi, 2, mean)

## Sepal.Length  Sepal.Width 
##        6.588        2.974

cVi = cov(grVi)
C = (nk - 1) * (cS + cVe + cVi)/(dim(iris)[1] - 3)
C

##              Sepal.Length Sepal.Width
## Sepal.Length   0.26500816  0.09272109
## Sepal.Width    0.09272109  0.11538776

eigen(C)

## eigen() decomposition
## $values
## [1] 0.30933556 0.07106036
## 
## $vectors
##            [,1]       [,2]
## [1,] -0.9022004  0.4313171
## [2,] -0.4313171 -0.9022004

set.seed(1)
train = sample(1:150, 75)

iris_train = iris[train, ]
iris_test = iris[-train, ]

iris_lda2 = lda(Species ~ Sepal.Length + Sepal.Width, data = iris_train, 
    prior = c(1, 1, 1)/3)

# Training error
table(predict(iris_lda2, newdata = iris_train)$class, iris_train$Species)

##             
##              setosa versicolor virginica
##   setosa         27          0         0
##   versicolor      1         15         8
##   virginica       0          5        19

# Test error
iris_lda2_predict = predict(iris_lda2, newdata = iris_test)
table(iris_lda2_predict$class, iris$Species[-train])

##             
##              setosa versicolor virginica
##   setosa         22          0         0
##   versicolor      0         22        11
##   virginica       0          8        12

Iris: training and test error for QDA

iris_qda2 = qda(Species ~ Sepal.Length + Sepal.Width, data = iris_train, 
    prior = c(1, 1, 1)/3)

Important: use the same division into training and test set for methods we want to compare.

# Training error
table(predict(iris_qda2, newdata = iris_train)$class, iris_train$Species)

##             
##              setosa versicolor virginica
##   setosa         28          0         0
##   versicolor      0         16         9
##   virginica       0          4        18

# Test error
iris_qda2_predict = predict(iris_qda2, newdata = iris_test)
table(iris_qda2_predict$class, iris$Species[-train])

##             
##              setosa versicolor virginica
##   setosa         22          0         0
##   versicolor      0         18        12
##   virginica       0         12        11

The QDA classifier has a training error rate of \(16\%\). When tested on a new set, the misclassification error rate was \[\text{Test error}_\text{QDA} = \frac{18}{75}=.24\]

The LDA classifier has given the smallest test error for classifying iris plants based on sepal width and sepal length for our test set and should be preferred in this case.

1. Would another division of the data into training and test set give the same conclusion (that LDA is better than QDA for this data set)? (A: Not necessarily, but probably.)

We will look into other choice than dividing into one training and one test set in Module 5 (crossvalidation).

2. What about the other two covariates? Would adding them to the model (4 covariates) give a better classification rule? (A: Probably. Try if you want.)

Fishers idea (Optional)

In 1936 R. A. Fisher developed LDA.

• His aim was to find a linear combination of the explanatory variables which maximized the ratio of its between class to within class variance.
• In this way the observations are transformed so that they are separated as much as possible.
• His approach has the advantage that it is suited for visual inspection and graphical description , it ``separates’’ the populations.

Let the between-class variance be denoted \(\boldsymbol{B}=\sum_{m=1}^{M}(\boldsymbol{\mu}_{m}- \bar{\boldsymbol{\mu}})(\boldsymbol{\mu}_{m}-\bar{\boldsymbol{\mu}})^{T}\), where \(\boldsymbol{\mu}_{m}\) denotes the mean of class \(\omega_{m}\) and \(\bar{\boldsymbol{\mu}}\) the overall mean.

The within-class variance is assumed equal for all classes and is denoted \(\boldsymbol{\Sigma}\) (\(\boldsymbol{\Sigma}\) is assumed to have full rank).

The linear combination \(\boldsymbol{l}^{T}\boldsymbol{X}\) that maximize \({\boldsymbol{l}^{T}\boldsymbol{B}\boldsymbol{l}}/{\boldsymbol{l}^{T}\boldsymbol{\Sigma}\boldsymbol{l}}\) under the constraint that \(\boldsymbol{l}^{T}\boldsymbol{\Sigma}\boldsymbol{l}=1\) is found to be the scaled eigenvectors of \(\boldsymbol{\Sigma}^{-1}\boldsymbol{B}\) corresponding to the nonzero eigenvalues of \(\boldsymbol{\Sigma}^{-1}\boldsymbol{B}\). The eigenvector corresponding to the largest eigenvalue defines the first discriminant \(\boldsymbol{l}_{1}^{T}\boldsymbol{X}\). The second linear discriminant \(\boldsymbol{l}_{2}^{T}\boldsymbol{X}\) is constructed from the eigenvector corresponding to the second largest eigenvalue and so on. (We also have \(Cov(\boldsymbol{l}_{j}^{T}\boldsymbol{X},\boldsymbol{l}_{i}^{T}\boldsymbol{X})=0\) for \(i \ne j\) and \(Var(\boldsymbol{l}_{j}^{T}\boldsymbol{X})=1\).)

The number of linear discriminants equals the number of nonzero eigenvalues. Observations are assigned to the class of the nearest (Euclidean distance) class mean in the discriminant space.

This equals classification to the nearest Mahalanobis distance population mean in the input space. Again, the parameters \(\boldsymbol{\mu}_{i}\) and \(\boldsymbol{\Sigma}\) and the between-class variance \(\boldsymbol{B}\) are usually unavailable. Replacing the parameters by estimates from the training set leads to Fisher’s sample linear discriminants.

Synthetic example for KNN classification

Consider a two-class example, and equal class prior probabilites.

A new observation \(x_0\) will be classified to \(A\) if \(P(Y=A | X=x_0) > 0.5\) and to class \(B\) otherwise.

• The figure below shows a plot of 100 observations from two classes \(A\) (red dots) and \(B\) (turquoise dots),
• simulated from a bivariate normal distribution with mean vectors \(\mu_A = (1, 1)^T\) and \(\mu_B = (3, 3)^T\) and a covariance matrix \(\Sigma_A = \Sigma_B = \begin{pmatrix} 2\hspace{2mm} 0 \\ 0 \hspace{2mm} 2 \end{pmatrix}\).
• We want to find a rule to classify a new observation to class \(A\) or \(B\).

Remark: since the truth is known here we can calculate the Bayes boundary and the Bayes error.

Since we have bivariate normal class distributions with common covariance matrix, the optimal boundary is given by LDA. The boundary will be at \(\delta_A({\bf x})=\delta_B({\bf x})\), where \(\delta_A({\bf x})= {\bf x}^T \boldsymbol{\Sigma}^{-1}\boldsymbol\mu_A - \frac{1}{2}\boldsymbol\mu_A^T \boldsymbol{\Sigma}^{-1}\boldsymbol\mu_A + \log \pi_A\), and for \(\delta_B({\bf x})\) with \(\boldsymbol\mu_B\).

\[{\bf x}^T \boldsymbol{\Sigma}^{-1}\boldsymbol\mu_A - \frac{1}{2}\boldsymbol\mu_A^T \boldsymbol{\Sigma}^{-1}\boldsymbol\mu_A + \log \pi_A={\bf x}^T \boldsymbol{\Sigma}^{-1}\boldsymbol\mu_B - \frac{1}{2}\boldsymbol\mu_B^T \boldsymbol{\Sigma}^{-1}\boldsymbol\mu_B + \log \pi_B\]

\[{\bf x}^T\boldsymbol{\Sigma}^{-1}(\boldsymbol\mu_A -\boldsymbol\mu_B)-\frac{1}{2}\boldsymbol\mu_A^T \boldsymbol{\Sigma}^{-1}\boldsymbol\mu_A +\frac{1}{2}\boldsymbol\mu_B^T \boldsymbol{\Sigma}^{-1}\boldsymbol\mu_B +\log \pi_A-\log \pi_B=0\]

Inserting numerical values gives: \(-x_1-x_2+4=0\), and boundary \(x_2=4-x_1\).

muA = matrix(c(1, 1), ncol = 1)
muB = matrix(c(3, 3), ncol = 1)
sigmainv = diag(2)/2
sigmainv %*% (muA - muB)

##      [,1]
## [1,]   -1
## [2,]   -1

-0.5 * t(muA) %*% sigmainv %*% muA + 0.5 * t(muB) %*% sigmainv %*% muB + 
    log(0.5) - log(0.5)

##      [,1]
## [1,]    4

The Bayes error can then be found by calculation of areas for the two class densities on the wrong side of the boundary, or by simulating many test data and counting misclassifications rates.

Synthetic data for KNN - continued

• Assume we have a new observation \(X_0 = (x_{01}, x_{02})^T\) which we want to classify as belonging to the class \(A\) or \(B\).
• To illustrate this problem we fit the \(K\)-nearest neighbor classifier to our simulated data set with \(K = 1, 3, 10\) and \(150\) and observe what happens.

In our plots, the small colored dots show the predicted classes for an evenly-spaced grid. The lines show the decision boundaries. If our new observation falls into the region within the red decision boundary, it will be classified as \(A\). If it falls into the region within the turqouise decision boundary, it will be classified as \(B\).

We see that the choice of \(K\) has a big influence on the result of our classification. By choosing \(K=1\) the classification is made to the same class as the one nearest neighbor. When \(K=3\) a majority vote is taken among the three nearest neighbors, and so on. We see that as \(K\) gets very large, the decision boundary tends towards a straight line (which is the Bayes boundary in this set-up).

To find the optimal value of \(K\) the typical procedure is to try different values of \(K\) and then test the predictive power of the different classifiers, for example by cross-validation, which will be discussed in Module 5.

We see that after trying all choices for \(K\) between 1 and 50, we see that a few choices of \(K\) gave the smallest misclassification error rate, estimating by leave-one out cross-validation (Leave-one-out cross-validation will be discussed in Module 5). The smallest error rate is equal to 0.165. This means that the classifier makes a misclassification 16.5% of the time and a correct classification 83.5% of the time.

This above example showed the bias-variance trade-off in a classification setting. Choosing a value of \(K\) amounts to choosing the correct level of flexibility of the classifier. This again is critical to the success of the classifier. A too low value of \(K\) will give a very flexible classifier (with high variance and low bias) which will fit the training set too well (it will overfit) and make poor predictions for new observations. Choosing a high value for \(K\) makes the classifier loose its flexibility and the classifier will have low variance but high bias.

The curse of dimensionality

The nearest neighbor classifier can be quite good if the number of predictor \(p\) is small and the number of observations \(n\) is large. We need enough close neighbors to make a good classification.

The effectiveness of the KNN classifier falls quickly when the dimension of the preditor space is high. This is because the nearest neighbors tend to be far away in high dimensions and the method no longer is local. This is referred to as the curse of dimensionality.

What to remember from Part A?

Using linear regression on a classification problem

Example 1: This example uses the Default data set from the ISLR package. Suppose we want to predict if a new customer will default or not based on his/her balance or income. We try to model this using a simple linear regression and a binary response variable: \[Y = \begin{cases} 1 \quad \text{ if default = "Yes"} \\ 0 \quad \text{ if default = "No"} \end{cases}.\] It would be tempting to do the classification according to the rule: classify as yes if \(\hat{Y}>0.5\), else as no.

The above plots shows default as a function of balance and default as a function of income with corresponding fitted linear regression lines (red for x=balance and orange for x=income). Notice that linear regression in this case produces predictions that are smaller than zero or bigger than one, this it is hard to interpret these as probabilities.

It is still possible to use linear regression for classification problems with two classes. It turns out that - if the conditional class densities are (multivariate) normal with equal covariance matrices then this linear regression (with 0 and 1 response) will in fact give the same classification as LDA. See e.g. Ripley (1995), Section 3.2.

Using linear regression on a classification problem?

Example 2: Suppose we want to classify a film. We have defined three classes: { drama, comedy, science-fiction}. We could try to model this using linear regression and the following coding: \[Y = \begin{cases} 1 \quad \text{if drama}, \\ 2 \quad \text{if comedy}, \\ 3 \quad \text{if science-fiction}.\end{cases}\] However, this coding implies an ordering of the variables and that the difference between the classes is equal. There is in general no natural way to code a quantitative variable with more than two classes such that it can be used with linear regression.

So, using linear regression to solve a classification problem seems hard with more than two classes - as done here. But, it turns out that using a dummy variable conding for the classes, it is possible to produce the same result as LDA (also with many classes). This is the starting point for flexible discriminant analysis.

Linear regression to do classification is not a bad idea, but requires some extra work (multivariate Y due to the dummy variable coding). Therefore we leave linear regression for now.

For two classes binary regression, in particular logistic regression, is very popular - and is up next.

The model

Assume that \(Y\) is coded (\(\mathcal{C} = \{1, 0\}\) or {success, failure}), and we focus on success. We may assume that \(Y_i\) follows a Bernoulli distribution with probability of success \(p_i\).

\[Y_i = \begin{cases} 1 \text{ with probability } p_i, \\ 0 \text{ with probability } 1-p_i. \end{cases}\]

In logistic regression we link together our covariates \({\bf x}_i\) with this probability \(p_i\) using a logistic function.

In the case of one covariate, the logistic function has the form: \[ p_i= \frac{e^{\beta_0+\beta_1 x_i}}{1 + e^{\beta_0 + \beta_1 x_i}}.\]

This function is S-shaped, and ranges between 0 and 1 (so the \(p_i\) is between 0 and 1). The parameter \(\beta_1\) determines the rate of increase or decrease of the S-shaped curve, and the sign indicates whether the curve ascends or descends.

Q: Where did that come from? There are other transforms that takes a linear predictor and transforms into the range 0 to 1.

Logistic regression ensures that the estimated probabilities lie in the interval between 0 and 1. This is illustrated in the figure below. The blue line shows the fitted line when performing logistic regression on default as a function of balance.

The parameters are estimated using the method of maximum likelihood - we will look at that soon, but first we look at how to interpret the estimated parameters.

library(ggplot2)
ggplot(data.frame(x = c(-6, 5)), aes(x)) + xlab(expression(x)) + ylab(expression(mu)) + 
    stat_function(fun = function(x) exp(x)/(1 + exp(x)), geom = "line", 
        colour = "red") + stat_function(fun = function(x) exp(2 * x)/(1 + 
    exp(2 * x)), geom = "line", colour = "orange") + stat_function(fun = function(x) exp(0.8 * 
    x)/(1 + exp(0.8 * x)), geom = "line", colour = "blue") + stat_function(fun = function(x) exp(1 + 
    x)/(1 + exp(1 + x)), geom = "line", colour = "red", linetype = "dashed") + 
    stat_function(fun = function(x) exp(1 + 2 * x)/(1 + exp(1 + 2 * x)), 
        geom = "line", colour = "orange", linetype = "dashed") + stat_function(fun = function(x) exp(1 + 
    0.8 * x)/(1 + exp(1 + 0.8 * x)), geom = "line", colour = "blue", 
    linetype = "dashed") + scale_colour_manual("0+k x", values = c("red", 
    "orange", "blue"), labels = c("1", "2", "0.8"))

The odds and the odds ratio

For the logistic regression it is hard to give a simple interpretation regression coefficients \(\beta\), because an increase in \(x_1\) by one unit does not give the same increase (decrease) in the probability \(p_i\) for all values of \(x_1\). But, looking at odds - it is simpler to explain what the \(\beta\)s mean.

For a probability \(p_i\) the ratio \(\frac{p_i}{1-p_1}\) is called the odds.

If \(p_i=\frac{1}{2}\) then the odds is \(1\), and if \(p_i=\frac{1}{4}\) then the odds is \(\frac{1}{3}\). We may make a table for probability vs. odds in R:

library(knitr)
library(kableExtra)
p = seq(0.1, 0.9, 0.1)
odds = p/(1 - p)
kable(t(data.frame(p, odds)), digits = c(2, 2), format = whichformat)

Odds may be seen to be a better scale than probability to represent chance, and is used in betting. In addition, odds are unbounded above.

Why is the odds relevant?

(Since \(p\) is used for probability we use \(r\) for number of covariates now.)

Let us assume that we have \(r\) covariates, and we use \(\eta_i\) (linear predictor) to help with our notation. \[\begin{align*} \eta_i&= \beta_0+\beta_1 x_{i1}+\beta_2 x_{i2}+\cdots + \beta_r x_{ir}\\ p_i&= \frac{\exp(\eta_i)}{1+\exp(\eta_i)}\\ \eta_i&=\ln(\frac{p_i}{1-p_i})\\ \ln(\frac{p_i}{1-p_i})&=\beta_0+\beta_1 x_{i1}+\beta_2 x_{i2}+\cdots + \beta_r x_{ir}\\ \frac{p_i}{1-p_i}=&\frac{P(Y_i=1|{\bf x}_i)}{P(Y_i=0|{\bf x}_i)}=\exp(\beta_0)\cdot \exp(\beta_1 x_{i1})\cdots\exp(\beta_r x_{ir}) \end{align*}\] We have a multiplicative model for the odds - which can help us to interpret our \(\beta\)s.

In addition we see that the logit of \(p_i\), \(\ln(\frac{p_i}{1-p_i})\), is linear in the \(\beta\)s (and in the \(x\)’s).

So, what if we increase \(x_{1i}\) to \(x_{1i}+1\)?

If the covariate \(x_{1i}\) increases by one unit (while all other covariates are kept fixed) then the odds is multiplied by \(\exp(\beta_1)\):

\[\begin{align*} \frac{P(Y_i=1\mid x_{i1}+1)}{P(Y_i=0)\mid x_{i1}+1)}&=\exp(\beta_0)\cdot \exp(\beta_1 (x_{i1}+1))\exp(\beta_2 (x_{i2}))\cdots\exp(\beta_r x_{ir})\\ &=\exp(\beta_0)\cdot \exp(\beta_1 x_{i1})\exp(\beta_1)\exp(\beta_2 x_{i2})\cdots\exp(\beta_r x_{ir})\\ &=\frac{P(Y_i=1\mid x_{i1})}{P(Y_i=0\mid x_{i1})}\cdot \exp(\beta_1)\\ \end{align*}\]

This means that if \(x_{i1}\) increases by \(1\) then: if \(\beta_1<0\) we get a decrease in the odds, if \(\beta_1=0\) no change, and if \(\beta_1>0\) we have an increase. Here \(\exp(\beta_1)\) is easier to interpret than \(\beta_1\).

colnames(Default)
fit = glm(default ~ student + balance + income, family = "binomial", 
    data = Default)
coef(fit)
round(exp(coef(fit)), 3)

## [1] "default" "student" "balance" "income" 
##   (Intercept)    studentYes       balance        income 
## -1.086905e+01 -6.467758e-01  5.736505e-03  3.033450e-06 
## (Intercept)  studentYes     balance      income 
##       0.000       0.524       1.006       1.000

Maximum likelihood estimation

We assume that pairs of covariates and responses \(\{x_i, y_i\}\) are measured independently of each other. Given \(n\) such observation pairs, the likelihood function of a logistic regression model can be written as: \[L(\boldsymbol{\beta}) = \prod_{i=1}^n L_i(\boldsymbol{\beta}) = \prod_{i=1}^n f(y_i; \boldsymbol{\beta}) = \prod_{i=1}^n (p_i)^{y_i}(1-p_i)^{1-y_i},\] where \(\boldsymbol{\beta} = (\beta_0, \beta_1, \beta_2, \ldots, \beta_r)^T\) enters into \(p_i\).

\[p_i= \frac{\exp(\beta_0+\beta_1 x_{i1}+\cdots + \beta_p x_{ir})}{1 + \exp(\beta_0 + \beta_1 x_{i1}+\cdots+\beta_r x_{ir})}\]

The maximum likelihood estimates are found by maximizing the likelihood, and since the log is a monotone transform (and maximizing the log-likelihood will give the same result as maximizing the likelihood) we usually work with the log-likelihood (because this makes the maths easier).

\[\begin{align*} \ln(L(\boldsymbol{\beta}))&=l(\boldsymbol{\beta}) =\sum_{i=1}^n \Big ( y_i \log p_i + (1-y_i) \log(1 - p_i )\Big ) \\ &= \sum_{i=1}^n \Big ( y_i \log \Big (\frac{p_i}{1-p_i} \Big) + \log(1-p_i) \Big ) \\ &= \sum_{i=1}^n \Big (y_i (\beta_0 + \beta_1 x_{i1}+\cdots + \beta_r x_{ir}) - \log(1 + e^{\beta_0 + \beta_1 x_{i1}+\cdots + \beta_p x_{ir}} \Big ).\end{align*}\]

• To maximize the log-likelihood function we find the \(r+1\) partial derivatives, and set equal til 0.
• This gives us a set of \(r+1\) non-linear equations in the \(\beta\)s.
• This set of equations does not have a closed form solution.
• These equations are therefore solved numerically. The Newton-Raphson algorithm (or Fisher Scoring) is used.

fit = glm(default ~ student + balance + income, family = "binomial", 
    data = Default)
summary(fit)

## 
## Call:
## glm(formula = default ~ student + balance + income, family = "binomial", 
##     data = Default)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.4691  -0.1418  -0.0557  -0.0203   3.7383  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.087e+01  4.923e-01 -22.080  < 2e-16 ***
## studentYes  -6.468e-01  2.363e-01  -2.738  0.00619 ** 
## balance      5.737e-03  2.319e-04  24.738  < 2e-16 ***
## income       3.033e-06  8.203e-06   0.370  0.71152    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 2920.6  on 9999  degrees of freedom
## Residual deviance: 1571.5  on 9996  degrees of freedom
## AIC: 1579.5
## 
## Number of Fisher Scoring iterations: 8

Inference

We may construct confidence intervals and test hypotheses about the \(\beta\)s, with the aim to understand which covariate that contributes to our posterior probabilites and classification.

This is done by assuming that each \(\hat{\beta}_j\) is approximately normally distributed with mean \(\beta_j\) and variance \(\hat{\text{Var}}(\hat{\beta}_j)\) (related to the negative of the inverse of the expected Hessian of the loglikelihood function).

The Akaike Information Criterion (AIC) for model selection

The AIC score is given by: \[AIC = 2 \cdot (r+1) -2 \cdot \text{loglik},\] where \(r+1\) is the number of model parameters. The loglik is the maximized log-likelihood \(l(\hat{\boldsymbol{\beta}})\) and \(\hat{\boldsymbol{\beta}}\) is the maximum-likelihood estimate of the parameter-vector \(\boldsymbol{\beta} = (\beta_0, \beta_1, ..., \beta_{r})^T\). The role of \(r+1\) is to penalize models with many parameters as a high number of parameters may lead to overfitting. The AIC value can be used to choose between candidate logistic regression models, where the model with the lowest AIC value is the one expected to give the best fit.

More about the AIC in Module 6.

Predictions

• We fit a (simple) logistic regression model to our data set, and
• get parameter estimates \(\hat{\beta}_0\) and \(\hat{\beta}_1\).
• We want to use this model to make a prediction when given a new observation \(x_0\).

\[\hat{p}(x_0) = \frac{e^{\hat{\beta}_0 + \hat{\beta}_1 x_0}}{1+e^{\hat{\beta}_0 + \hat{\beta}_1 x_0}} \]

This \(\hat{p}(x_0)\) is the estimated probability that the new observation \(x_0\) belongs to the class defined by \(Y=1\).

In the case of qualitative covariates, a dummy variable needs to be introduced. This is done in a similar fashion as for linear regression.

Want to learn more (theory) about logistic regression?

In TMA4315 Generalized linear models we spent 3 weeks with binary regression - mainly logistic regression. The focus there was on all parts of the regression (not classification) with a mathematical focus on estimation, inference, model fit.

Example: South African heart disease data set

In this example we use the SAhert data set from the ElemStatLearn package. This is a retrospective sample of males in a heart-disease high-risk region in South Africa. It consists of 462 observations on the 10 variables. All subjects are male in the age range 15-64. There are 160 cases (individuals who have suffered from a conorary heart disease) and 302 controls (individuals who have not suffered from a conorary heart disease).

The response value (chd) and covariates

• chd : conorary heart disease {yes, no} coded by the numbers {1, 0}
• sbp : systolic blood pressure
  
• tobacco : cumulative tobacco (kg)
  
• ldl : low density lipoprotein cholesterol
• adiposity : a numeric vector
• famhist : family history of heart disease. Categorical variable with two levels: {Absent, Present}.
• typea : type-A behavior
• obesity : a numerical value
• alcohol : current alcohol consumption
• age : age at onset

The goal is to identify important risk factors. We start by loading and looking at the data:

library(ElemStatLearn)
library(knitr)
library(kableExtra)
heartds = SAheart
heartds$chd = as.factor(heartds$chd)
kable(head(heartds), format = whichformat)

In order to investigate the data further, we use the ggpairs function from the GGally library, to make scatter plots of the covariates. The coloring is done according to the response variable, where green represents a case \(Y=1\) and red represents a control \(Y=0\).

library(ggplot2)
library(GGally)
ggpairs(heartds, ggplot2::aes(color=chd), #upper="blank",  
        lower = list(continuous = wrap("points", alpha = 0.3, size=0.2)))

We now fit a (multiple) logistic regression model using the glm function and the full data set. In order to fit a logistic model, the family argument must be set equal to ="binomial". The summary function prints out the estimates of the coefficients, their standard errors and z-values. As for a linear regression model, the significant coefficients are indicated by stars where the significant codes are included in the R outprint.

glm_heart = glm(chd ~ ., data = heartds, family = "binomial")
summary(glm_heart)

## 
## Call:
## glm(formula = chd ~ ., family = "binomial", data = heartds)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -1.7781  -0.8213  -0.4387   0.8889   2.5435  
## 
## Coefficients:
##                  Estimate Std. Error z value Pr(>|z|)    
## (Intercept)    -6.1507209  1.3082600  -4.701 2.58e-06 ***
## sbp             0.0065040  0.0057304   1.135 0.256374    
## tobacco         0.0793764  0.0266028   2.984 0.002847 ** 
## ldl             0.1739239  0.0596617   2.915 0.003555 ** 
## adiposity       0.0185866  0.0292894   0.635 0.525700    
## famhistPresent  0.9253704  0.2278940   4.061 4.90e-05 ***
## typea           0.0395950  0.0123202   3.214 0.001310 ** 
## obesity        -0.0629099  0.0442477  -1.422 0.155095    
## alcohol         0.0001217  0.0044832   0.027 0.978350    
## age             0.0452253  0.0121298   3.728 0.000193 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 596.11  on 461  degrees of freedom
## Residual deviance: 472.14  on 452  degrees of freedom
## AIC: 492.14
## 
## Number of Fisher Scoring iterations: 5

How did we find the coefficients and what does the second column mean?

Multinomial logistic regression

The logistic regression model can be generalized for a response variable with more than two classes. Assume we have a response variable with \(K\) possible classes and \(r\) covariates. The probability that \(Y\) belongs to class \(k\), given an observation vector \(\mathbf{x} = (x_1, x_2, \dots, x_r)^T\) is (usually) modelled by:

\[\ln \frac{P(Y = k | \mathbf{x})}{P(Y = K | \mathbf{x})}= \beta_{0k} + \beta_{1k} x_1 + \cdots + \beta_{rk} x_r.\]

The multinomial logistic regression model is implemented in the glmnet or VGAM package in R.

We will not discuss this further since LDA is more popular (than logistic regression) in the multi-class setting. And, as we shall see soon - they are not that different.

Example Continued: South African heart disease

We want to evaluate our multiple logistic model for the SAheart data set. In order to investigate the training error and the test error, we divide the original data set, randomly, into two samples of equal size.

set.seed(20)
train_ID = sample(1:nrow(heartds), nrow(heartds)/2)
train_SA = heartds[train_ID, ]
test_SA = heartds[-train_ID, ]

We now fit a logistic regression model, using the training set only:

glm_SA = glm(chd ~ ., data = train_SA, family = "binomial")
summary(glm_SA)

## 
## Call:
## glm(formula = chd ~ ., family = "binomial", data = train_SA)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.0583  -0.7660  -0.4523   0.8679   2.2682  
## 
## Coefficients:
##                  Estimate Std. Error z value Pr(>|z|)   
## (Intercept)    -5.4999981  1.9010122  -2.893  0.00381 **
## sbp             0.0050738  0.0089677   0.566  0.57154   
## tobacco         0.1228036  0.0443098   2.771  0.00558 **
## ldl             0.0315840  0.0817017   0.387  0.69907   
## adiposity       0.0223541  0.0398203   0.561  0.57454   
## famhistPresent  0.7763880  0.3434487   2.261  0.02379 * 
## typea           0.0328267  0.0181739   1.806  0.07088 . 
## obesity        -0.0487256  0.0592647  -0.822  0.41098   
## alcohol        -0.0007868  0.0079569  -0.099  0.92123   
## age             0.0470105  0.0170190   2.762  0.00574 **
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 289.73  on 230  degrees of freedom
## Residual deviance: 229.18  on 221  degrees of freedom
## AIC: 249.18
## 
## Number of Fisher Scoring iterations: 5

By comparing this outprint with the corresponding outprint above, we see that the estimated coefficients slightly differ. This is because a different data set has been used to fit the model. We previously used the full data set.

We want to estimate the probability of a chd event for the observations in the test set. To do this we can insert the estimated coefficient into the logistic equation, remembering that famhist is a categorical covariate, modeled by a dummy variable: \[x_\text{famhist} = \begin{cases} 1, \quad \text{ if Present}, \\ 0, \quad \text{ if Absent}.\end{cases}\] The estimated probability of \(Y=1\) if famhist = "Present", given a vector \(\mathbf{X}\) of covariate observations is: \[\hat{p}(\mathbf{X}) =\frac{e^{-7.43 + 0.01 x_{\text{sbp}} + 0.09 x_{\text{tobacco}} + 0.16 x_\text{ldh} + 0.01 x_\text{adiposity}+1.04 \cdot 1 + 0.04 x_\text{typea}-0.04 x_\text{obesity}-0.005 x_\text{alcohol} +0.05 x_{\text{age}}}}{1+e^{-7.43 + 0.01 x_{\text{sbp}} + 0.09 x_{\text{tobacco}} + 0.16 x_\text{ldh} + 0.01 x_\text{adiposity}+1.04 \cdot 1 + 0.04 x_\text{typea}-0.04 x_\text{obesity}-0.005 x_\text{alcohol} +0.05 x_{\text{age}}}} .\] Whereas, if famhist = "Absent" the estimated probability is: \[\hat{p}(\mathbf{X}) =\frac{e^{-7.43 + 0.01 x_{\text{sbp}} + 0.09 x_{\text{tobacco}} + 0.16 x_\text{ldh} + 0.01 x_\text{adiposity}+1.04 \cdot 0 + 0.04 x_\text{typea}-0.04 x_\text{obesity}-0.005 x_\text{alcohol} +0.05 x_{\text{age}}}}{1+e^{-7.43 + 0.01 x_{\text{sbp}} + 0.09 x_{\text{tobacco}} + 0.16 x_\text{ldh} + 0.01 x_\text{adiposity}+1.04 \cdot 0 + 0.04 x_\text{typea}-0.04 x_\text{obesity}-0.005 x_\text{alcohol} +0.05 x_{\text{age}}}} .\]

The predict function does these calculations for us. When specifying type="response" the function returns the probabilities for \(Y=1\).

probs_SA = predict(glm_SA, newdata = test_SA, type = "response")

From these probabilities we can obtain classifications, by specifying a threshold value. We have here chosen a threshold value of 0.5. By using the ifelse function we specify that all probabilities larger than 0.5 are to be classified as 1, while the remaining probabilities are to be classified as 0.

pred_SA = ifelse(probs_SA > 0.5, 1, 0)

predictions_SA = data.frame(probs_SA, pred_SA, test_SA[, 10])
colnames(predictions_SA) = c("Estim. prob. of Y=1", "Predicted class", 
    "True class")
kable(head(predictions_SA), format = whichformat, booktabs = TRUE)

We can now use the confusion matrix to count the number of misclassifications. The below confusion matrix is calculated using the test set and comparing the predicted classes with the true classes.

table(pred_SA, SAheart[-train_ID, 10])

##        
## pred_SA   0   1
##       0 117  44
##       1  28  42

The logistic model has correctly classified 130+40 times, and misclassified 24+37 times. The misclassification test error rate is thus: \[\text{Test error} = \frac{24+37}{231} \approx 0.264\]

The training error can be calculated in a similar fashion, but now we use the fitted model to make prediction for the training set.

SA_train_prob = glm_SA$fitted.values
SA_train_pred = ifelse(SA_train_prob > 0.5, 1, 0)
conf_train = table(SA_train_pred, SAheart[train_ID, 10])
misclas_train = (231 - sum(diag(conf_train)))/231
misclas_train

## [1] 0.2380952

The train misclassification error rate is \(\approx 25.1\%\).

ROC curves and AUC

The receiver operating characteristics (ROC) curve gives a graphical display of the sensitivity against specificity, as the threshold value (cut-off on probability of success or disease) is moved over the range of all possible values. An ideal classifier will give a ROC curve which hugs the top left corner, while a straight line represents a classifier with a random guess of the outcome.

The AUC score is the area under the AUC curve. It ranges between the values 0 and 1, where a higher value indicates a better classifier. The AUC score is useful for comparing the performance of different classifiers, as all possible threshold values are taken into account.

library(pROC)
SA_roc = roc(SAheart[-train_ID, 10], probs_SA, legacy.axes = TRUE)
ggroc(SA_roc) + ggtitle("ROC curve") + annotate("text", x = 0.25, y = 0.3, 
    label = "AUC = 0.7762")

res = table(pred_SA, SAheart[-train_ID, 10])
sens = res[2, 2]/sum(res[2, ])
spec = res[1, 1]/sum(res[1, ])
sens

## [1] 0.6

spec

## [1] 0.7267081

Advantages of discriminant analysis

• Discriminant analysis is more stable than logistic regression when the classes are well-separated.
• Discriminant analysis is more stable than logistic regression if the number of observations \(n\) is small and the distribution of the predictors \({\bf X}\) is approximately (multivariate) normal.

Linearity

Assume a binary classification problem with one covariate. Recall that logistic regression can be written: \[\log \Big ( \frac{p(x)}{1-p(x)}\Big ) = \beta_0 + \beta_1 x.\]

For LDA we have that \(p_0(x)\) is the probability that the observation \(x\) belongs to class 0, while \(p_1(x)=1-p_0(x)\) is the probability that it belongs to class 1.

Observe that this show that our class boundary is linear.

Compulsory Exercise 1, Problem 3a.

\[\begin{align*}\log \frac{P(Y=0 | X = x)}{P(Y=1 | X=x)} &= \log \frac{\pi_0}{\pi_1} + \log \frac{f_0(x)}{f_1(x)} \\ &= \log \frac{\pi_0}{\pi_1} -\frac{1}{2 \sigma^2}(x-\mu_0)^2 + \frac{1}{2 \sigma^2}(x-\mu_1)^2 \\ &= \log \frac{\pi_0}{\pi_1} - \frac{1}{2 \sigma^2} (x^2 - 2 x \mu_0 + \mu_0^2 - x^2 + 2 x \mu_1 - \mu_1^2) \\ &= \log \frac{\pi_0}{\pi_1} -\frac{1}{2 \sigma^2}(\mu_0^2 - \mu_1^2)+\frac{1}{\sigma^2}(\mu_0 - \mu_1) x \\ &= \alpha_0 + \alpha_1 x \end{align*}\]

The two methods can thus be written in the same form (linear in parameters and in \(x\)). The difference is in how the parameters \((\alpha_0,\alpha_1,\beta_0,\beta_1)\) are estimated.

Extensions for classifications

• Module 5: how to use cross-validation in model evaluation and model selection
• (Module 6: model selection - but mainly regression)
• Module 7: maybe a taste of nonlinear methods
• Module 8: classification trees (binary splits for the covariates)
• Module 9: support vector machines
• Module 11: neural nets

# Recommended exercises

Theoretical exercises

Data analysis with R

Compulsory exercise 1 in 2018: Problem - Classification

In this problem, we use a wine dataset of chemical measurement of two variables, Color_intensity and Alcalinity_of_ash, on 130 wines from two cultivars in a region in Italy.

The data set is a subset of a data set from https://archive.ics.uci.edu/ml/datasets/Wine, see that page for information of the source of the data.

Below you find code to read the data, plot the data and to divide the data into a training set and a test set. To get your own unique division please change the seed (where it says set.seed(4268) you change 4268 to your favorite number).

library(ggplot2)
library(GGally)
library(class)
library(MASS)
library(pROC)
wine = read.csv("https://www.math.ntnu.no/emner/TMA4268/2018v/data/Comp1Wine.csv", 
    sep = " ")
wine$class = as.factor(wine$class - 1)
colnames(wine) = c("y", "x1", "x2")
ggpairs(wine, ggplot2::aes(color = y))

n = dim(wine)[1]
set.seed(4268)  #to get the same order if you rerun
ord = sample(1:n)  #shuffle, without replacement 
test = wine[ord[1:(n/2)], ]  # first half is test, second train - could have been opposite
train = wine[ord[((n/2) + 1):n], ]

In our data the two classes are named y and coded \(Y=0\) and \(Y=1\), and we name \(x_1\)= Alcalinity_of_ash=x1 and \(x_2\)=Color_intensity=x2.

KNN3 = knn(train = train[, -1], test = test[, -1], k = 3, cl = train$y, 
    prob = F)
t3 = table(test$y, KNN3)
t3
apply(t3, 1, sum)
n = length(test$y)
error = (n - sum(diag(t3)))/n
error
KNN3probwinning = attributes(knn(train = train[, -1], test = test[, -1], 
    k = 3, cl = train$y, prob = TRUE))$prob
KNN3prob <- ifelse(KNN3 == "0", 1 - KNN3probwinning, KNN3probwinning)
# cbind(KNN3prob,KNN3,KNN3probwinning) to check that this is correct
KNN3roc = roc(response = test$y, predictor = KNN3prob)
cbind(KNN3roc$threshold, KNN3roc$sens, KNN3roc$specificities)

KNN9 = knn(train = train[, -1], test = test[, -1], k = 9, cl = train$y, 
    prob = F)
t9 = table(test$y, KNN9)
t9
error = (n - sum(diag(t9)))/n
error
KNN9probwinning = attributes(knn(train = train[, -1], test = test[, -1], 
    k = 9, cl = train$y, prob = TRUE))$prob
KNN9prob <- ifelse(KNN9 == "0", 1 - KNN9probwinning, KNN9probwinning)
KNN9roc = roc(response = test$y, predictor = KNN9prob)
cbind(KNN9roc$threshold, KNN9roc$sens, KNN9roc$specificities)
table(KNN3, KNN9)

Exam 2018 Problem 3: Linear discriminant analysis

In this problem you may use that the probability density function for a \(p\)-dimensional multivariate normal random variable \({\bf X}\) with mean \({\boldsymbol \mu}\) and variance-covariance matrix \({\boldsymbol \Sigma}\) is given as: \[f({\bf x}) = \frac{1}{(2 \pi)^{p/2}|\boldsymbol{\Sigma}|^{1/2}}\exp({-\frac{1}{2}({\bf x}-\boldsymbol\mu)^T \boldsymbol{\Sigma}^{-1}({\bf x}-\boldsymbol\mu)})\] We will not discuss how to estimate the mean and variance-covariance matrix here, so you may assume that they are known.

Q10: Write down the mathematical model assumptions for a linear discriminant analysis with two classes (coded as 0 and 1) and \(p\) explanatory variables and explain what the different ingredients are.

Q11: Explain how you derive the mathematical formula for the posterior probability for class 1.

Q12: Derive the mathematical formula for the class boundary between the two classes, given that the classification rule is to classify to the class with the highest posterior probability.

Q13: Is this boundary linear or non-linear in the space of the explanatory variables?

Exam 2018 Problem 4ab: Classification of diabetes cases

We will look at data on diabetes (diabetes is 0 if not present and 1 if present) from a population of women of Pima Indian heritage in the US, available in the R MASS package. The following covariates were collected for each woman:

• npreg: number of pregnancies
• glu: plasma glucose concentration in an oral glucose tolerance test
• bp: diastolic blood pressure (mmHg)
• skin: triceps skin fold thickness (mm)
• bmi: body mass index (weight in kg/(height in m)\(^2\))
• ped: diabetes pedigree function.
• age: age in years

We will use a training set (called train) with 200 observations (132 non-diabetes and 68 diabetes cases) and a test set (called test) with 332 observations (223 non-diabetes and 109 diabetes cases). Our aim is to make a classification rule for diabetes (or not) based on the available data.

Below you find R-code and results from fitting a logistic regression to the train data set.

Q14: Write down the statistical model for the logistic regression.

Q15: Explain what is the estimated effect of the ped covariate on getting diabetes.

Q16: Would you predict that a person with the following characteristics has diabetes?

Person: npreg=2, glu=145, bp=85, skin=35, bmi=37, ped=0.7, age=40.

Q17: Draw a feedforward neural network with the same architecture as the logistic regression and specify which activation function(s) is/are used. Label nodes and connections with the same notation as for the mathematical model for the logistic regression.

The test data set was used to evaluate the performance of the logistic regression. A receiver–operator curve (ROC) was constructed and is the solid curve in the figure below (the dotted and dashed curves will be studie in Q24).

Q18: Explain briefly how a ROC curve is constructed. Your explanation should include the words cut-off, confusion matrix, sensitivity, specificity.

Look at the confusion matrix (probability cut-off 0.5) reported for the test set in the bottom part of the print-out in the figure below.

Q19: Which point on the ROC curve does this cut-off correspond to?

# logistic regression
> fitlogist=glm(diabetes~npreg+glu+bp+skin+bmi+ped+age,data=train,
family=binomial(link="logit"))
> summary(fitlogist)
Call:
glm(formula = diabetes ~ npreg + glu + bp + skin + bmi + ped + 
    age, family = binomial(link = "logit"), data = train)
Coefficients:
             Estimate Std. Error z value Pr(>|z|)    
(Intercept) -9.773062   1.770386  -5.520 3.38e-08 ***
npreg        0.103183   0.064694   1.595  0.11073    
glu          0.032117   0.006787   4.732 2.22e-06 ***
bp          -0.004768   0.018541  -0.257  0.79707    
skin        -0.001917   0.022500  -0.085  0.93211    
bmi          0.083624   0.042827   1.953  0.05087 .  
ped          1.820410   0.665514   2.735  0.00623 ** 
age          0.041184   0.022091   1.864  0.06228 .  
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
(Dispersion parameter for binomial family taken to be 1)
    Null deviance: 256.41  on 199  degrees of freedom
Residual deviance: 178.39  on 192  degrees of freedom
AIC: 194.39
Number of Fisher Scoring iterations: 5

> predlogist=predict(fitlogist,newdata=test,type="response")
> testclasslogist=ifelse(predlogist > 0.5, 1, 0)
> table(test$diabetes, testclasslogist)
   testclasslogist
      0   1
  0 200  23
  1  43  66
>  # note: 223 true non-diabetes and 109 true diabetes cases
> library(pROC)
> roclogist=roc(test$diabetes,predlogist)
> auc(roclogist)
Area under the curve: 0.8659
> plot(roclogist, lty="solid")

ROC-curves for logistic (solid), bagged trees (dashed) and quadratic discriminant analysis (dotted) for the diabetes test data set.